What is the integral of (sin(2e^{-4x}))/(e^{4x)} ?

The integral of (sin(2e^{-4x}))/(e^{4x)} is 1/8 cos(2e^{-4x})+C

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integral of (sin(2e^{-4x}))/(e^{4x)}

\int \frac{sin ( 2 e ^{- 4 x} )}{e ^{4 x}} d x

\frac{1}{8} cos (2 e^{- 4 x}) + C

Solution steps

\int \frac{sin ( 2 e ^{- 4 x} )}{e ^{4 x}} d x

Apply u-substitution

: \int - \frac{sin ( u )}{8} d u

= \int - \frac{sin ( u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{8} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= - \frac{1}{8} (- cos (u))

Substitute back

u = 2 e^{- 4 x}

= - \frac{1}{8} (- cos (2 e^{- 4 x}))

Simplify

= \frac{1}{8} cos (2 e^{- 4 x})

Add a constant to the solution

= \frac{1}{8} cos (2 e^{- 4 x}) + C

f (x) = x^{3} - 3 x^{2} + 3 x, g (x) = x

f (3) = 2 x^{2} + 4 x

\int \frac{1}{x ^{2} 25 x ^{2} + 64} d x

\frac{\partial}{\partial y} (- 3 x^{2} y)

f (x) = x^{2} + 4 x, g (x) = x + 10

What is the integral of (sin(2e^{-4x}))/(e^{4x)} ?
The integral of (sin(2e^{-4x}))/(e^{4x)} is 1/8 cos(2e^{-4x})+C